movie question.

$\text{circle}=\text{Range}[100];\text{While}[\text{Length}[\text{circle}]>1,\text{circle}=\text{Drop}[\text{RotateLeft}[\text{circle}],1]];\text{circle}\Rightarrow \{73\}$

The following English description, using Google Translate, translates using to both Hindi and Arabic and then return translation to English.

Make a list of 100 numbers.

Repeat the following process until there is only one number remaining: move the first number to the other end of the list (the person lives) and then remove the first number of the list (the person dies).

Show last number.

First define pair $(N,S) \in \{1,\dots,100\} \times \{1,-1\}$ , where the first element is the number of the remaining individuals, after each round of slay, and $S$ indicates the first person in the queue who gets the sword. $S=1$ means the first person in the queue gets the sword and $S=-1$ means the second person in the queue gets the sword.

Also we define the parity of $N$ as $p(N)=1$ if the parity is even and $p(N)=-1$ when the parity is odd.

Now we set a rule. Given $(N,S)(t)$ , $(N,S)(t+1)$ is given by

$N(t+1) = \left\{\begin{array}{lr} \frac{N(t)}{2} & if p(N(t))=1, S(t)=1\\ \lfloor \frac{N(t)}{2} \rfloor & if p(N(t))=-1,S(t)=-1\\ \lceil \frac{N(t)}{2} \rceil & if p(N(t))=-1,S(t)=1\\ \end{array}\right\}$

$S(t+1) = S(t)\times p(N(t))$

using the rules and pen and paper, we get the following sequence pairs.

$(100,1),(50,1)(25,1)(13,-1)(6,1)(3,1)(2,-1)(1,-1)$

every time we have a $1$ as the starting point, we map a number $x \in \{1,2,\dots,100\}$ to $\frac{x+1}{2}$ and when we have a $-1$ as the starting point, we map $x \in \{1,2,\dots,100\}$ to $\frac{x}{2}$ . mathematically

$\frac{\frac{\frac{\frac{\frac{x+1}{2}+1}{2}+1}{2^2}+1}{2}+1}{2^2}$

The individual who would survive would have a number $x$ S.T. when it is passed to the map above, the final result would be $1$ . So, we set it equal to $1$ and solve it for $x$ . the result is $x=73$